Why the homomorphism from g acting on a to left coset of stabilizer of a is surjective?

Question

Suppose $b = g \cdot a$. Then $gG_a$ is the left coset of $G_a$. The map $b = g \cdot a \rightarrow gG_a$ is a map from $C_a$ to the set of left cosets of $G_a$ in $G$. Dummit says this map is surjective because for any $g \in G$ the element $g \cdot a$ is an element of $C_a$. But I don't quite get it ....Plenty of thanks!

Answer 1

$G_a=\{g \in G: g \cdot a=a\}$ ($G$-stabilizer of $a$) and $C_a=\{g \cdot a: g \in G\}$ ($G$-orbit of $a$). Define a map $$f : C_a \rightarrow L:=\{gG_a: g \in G\}$$ by $f(g \cdot a)=gG_a$. This map is surjective, for let $gG_a \in L$, then obviously $f(g \cdot a)=gG_a$. But in addition one has to show that the element $g \cdot a$ is independent of the left-coset representative $g$. So assume that $gG_a=hG_a$. Then $h^{-1}g \in G_a$, so $(h^{-1}g) \cdot a=a$, implying $g \cdot a= h \cdot a$. And then we are done! In the same way you show $f$ is injective. Hence $\#C_a=\text {index}[G:G_a]$.